Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer:

(4, 5), (7, 6), (4, 3), (1, 2)

Let coordinates be A(4, 5), B(7, 6), C(4, 3) and D(1, 2).

By formula,

Distance between two points (D) = $\sqrt{(x$2 - x1)^2 + (y2 - y1)^2}

Calculating sides, we get :

$AB = \sqrt{(7 - 4)^2 + [6 - 5]^2} \\[1em] = \sqrt{(3)^2 + (1)^2} \\[1em] = \sqrt{9 + 1} \\[1em] = \sqrt{10} \text{ units} \\[1em] BC = \sqrt{(4 - 7)^2 + (3 - 6)^2} \\[1em] = \sqrt{(-3)^2 + (-3)^2} \\[1em] = \sqrt{9 + 9} \\[1em] = \sqrt{18} \text{ units} \\[1em] CD = \sqrt{(1 - 4)^2 + (2 - 3)^2} \\[1em] = \sqrt{(-3)^2 + (-1)^2} \\[1em] = \sqrt{9 + 1} \\[1em] = \sqrt{10} \text{ units} \\[1em] AD = \sqrt{(1 - 4)^2 + (2 - 5)^2} \\[1em] = \sqrt{(-3)^2 + (-3)^2} \\[1em] = \sqrt{9 + 9} \\[1em] = \sqrt{18} \text{ units} \\[1em]$

Calculating diagonals, we get :

$AC = \sqrt{(4 - 4)^2 + (3 - 5)^2} \\[1em] = \sqrt{0^2 + (-2)^2} \\[1em] = \sqrt{0 + 4} \\[1em] = 2 \text{ units}. \\[1em] BD = \sqrt{(1 - 7)^2 + (2 - 6)^2} \\[1em] = \sqrt{(-6)^2 + (-4)^2} \\[1em] = \sqrt{36 + 16} \\[1em] = \sqrt{52} \text{ units}.$

Since, AB = CD and BC = AD and AC ≠ BD.

Hence, the above points form a parallelogram.